\(\int \frac {x (c+d x^2)^{3/2}}{(a+b x^2)^2} \, dx\) [743]

Optimal result

Integrand size = 22, antiderivative size = 99 \[ \int \frac {x \left (c+d x^2\right )^{3/2}}{\left (a+b x^2\right )^2} \, dx=\frac {3 d \sqrt {c+d x^2}}{2 b^2}-\frac {\left (c+d x^2\right )^{3/2}}{2 b \left (a+b x^2\right )}-\frac {3 d \sqrt {b c-a d} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{2 b^{5/2}} \]

[Out]

-1/2*(d*x^2+c)^(3/2)/b/(b*x^2+a)-3/2*d*arctanh(b^(1/2)*(d*x^2+c)^(1/2)/(-a*d+b*c)^(1/2))*(-a*d+b*c)^(1/2)/b^(5
/2)+3/2*d*(d*x^2+c)^(1/2)/b^2

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {455, 43, 52, 65, 214} \[ \int \frac {x \left (c+d x^2\right )^{3/2}}{\left (a+b x^2\right )^2} \, dx=-\frac {3 d \sqrt {b c-a d} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{2 b^{5/2}}-\frac {\left (c+d x^2\right )^{3/2}}{2 b \left (a+b x^2\right )}+\frac {3 d \sqrt {c+d x^2}}{2 b^2} \]

[In]

Int[(x*(c + d*x^2)^(3/2))/(a + b*x^2)^2,x]

[Out]

(3*d*Sqrt[c + d*x^2])/(2*b^2) - (c + d*x^2)^(3/2)/(2*b*(a + b*x^2)) - (3*d*Sqrt[b*c - a*d]*ArcTanh[(Sqrt[b]*Sq
rt[c + d*x^2])/Sqrt[b*c - a*d]])/(2*b^(5/2))

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + 1))), x] - Dist[d*(n/(b*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 455

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m
- n + 1, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {(c+d x)^{3/2}}{(a+b x)^2} \, dx,x,x^2\right ) \\ & = -\frac {\left (c+d x^2\right )^{3/2}}{2 b \left (a+b x^2\right )}+\frac {(3 d) \text {Subst}\left (\int \frac {\sqrt {c+d x}}{a+b x} \, dx,x,x^2\right )}{4 b} \\ & = \frac {3 d \sqrt {c+d x^2}}{2 b^2}-\frac {\left (c+d x^2\right )^{3/2}}{2 b \left (a+b x^2\right )}+\frac {(3 d (b c-a d)) \text {Subst}\left (\int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx,x,x^2\right )}{4 b^2} \\ & = \frac {3 d \sqrt {c+d x^2}}{2 b^2}-\frac {\left (c+d x^2\right )^{3/2}}{2 b \left (a+b x^2\right )}+\frac {(3 (b c-a d)) \text {Subst}\left (\int \frac {1}{a-\frac {b c}{d}+\frac {b x^2}{d}} \, dx,x,\sqrt {c+d x^2}\right )}{2 b^2} \\ & = \frac {3 d \sqrt {c+d x^2}}{2 b^2}-\frac {\left (c+d x^2\right )^{3/2}}{2 b \left (a+b x^2\right )}-\frac {3 d \sqrt {b c-a d} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{2 b^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.97 \[ \int \frac {x \left (c+d x^2\right )^{3/2}}{\left (a+b x^2\right )^2} \, dx=\frac {\sqrt {c+d x^2} \left (-b c+3 a d+2 b d x^2\right )}{2 b^2 \left (a+b x^2\right )}-\frac {3 d \sqrt {-b c+a d} \arctan \left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {-b c+a d}}\right )}{2 b^{5/2}} \]

[In]

Integrate[(x*(c + d*x^2)^(3/2))/(a + b*x^2)^2,x]

[Out]

(Sqrt[c + d*x^2]*(-(b*c) + 3*a*d + 2*b*d*x^2))/(2*b^2*(a + b*x^2)) - (3*d*Sqrt[-(b*c) + a*d]*ArcTan[(Sqrt[b]*S
qrt[c + d*x^2])/Sqrt[-(b*c) + a*d]])/(2*b^(5/2))

Maple [A] (verified)

Time = 3.06 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.10

method	result	size
pseudoelliptic	\(\frac {-\frac {3 d \left (b \,x^{2}+a \right ) \left (a d -b c \right ) \arctan \left (\frac {b \sqrt {d \,x^{2}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{2}+\frac {3 \sqrt {\left (a d -b c \right ) b}\, \sqrt {d \,x^{2}+c}\, \left (\frac {\left (2 d \,x^{2}-c \right ) b}{3}+a d \right )}{2}}{b^{2} \left (b \,x^{2}+a \right ) \sqrt {\left (a d -b c \right ) b}}\)	\(109\)
risch	\(\frac {d \sqrt {d \,x^{2}+c}}{b^{2}}-\frac {-\frac {\left (a d -b c \right ) d \ln \left (\frac {-\frac {2 \left (a d -b c \right )}{b}-\frac {2 d \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {d \left (x +\frac {\sqrt {-a b}}{b}\right )^{2}-\frac {2 d \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}}{x +\frac {\sqrt {-a b}}{b}}\right )}{b \sqrt {-\frac {a d -b c}{b}}}-\frac {\left (a d -b c \right ) d \ln \left (\frac {-\frac {2 \left (a d -b c \right )}{b}+\frac {2 d \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {d \left (x -\frac {\sqrt {-a b}}{b}\right )^{2}+\frac {2 d \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}}{x -\frac {\sqrt {-a b}}{b}}\right )}{b \sqrt {-\frac {a d -b c}{b}}}-\frac {\sqrt {-a b}\, \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \left (\frac {b \sqrt {d \left (x +\frac {\sqrt {-a b}}{b}\right )^{2}-\frac {2 d \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}}{\left (a d -b c \right ) \left (x +\frac {\sqrt {-a b}}{b}\right )}+\frac {d \sqrt {-a b}\, \ln \left (\frac {-\frac {2 \left (a d -b c \right )}{b}-\frac {2 d \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {d \left (x +\frac {\sqrt {-a b}}{b}\right )^{2}-\frac {2 d \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}}{x +\frac {\sqrt {-a b}}{b}}\right )}{\left (a d -b c \right ) \sqrt {-\frac {a d -b c}{b}}}\right )}{4 a \,b^{2}}+\frac {\sqrt {-a b}\, \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \left (\frac {b \sqrt {d \left (x -\frac {\sqrt {-a b}}{b}\right )^{2}+\frac {2 d \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}}{\left (a d -b c \right ) \left (x -\frac {\sqrt {-a b}}{b}\right )}-\frac {d \sqrt {-a b}\, \ln \left (\frac {-\frac {2 \left (a d -b c \right )}{b}+\frac {2 d \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {d \left (x -\frac {\sqrt {-a b}}{b}\right )^{2}+\frac {2 d \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}}{x -\frac {\sqrt {-a b}}{b}}\right )}{\left (a d -b c \right ) \sqrt {-\frac {a d -b c}{b}}}\right )}{4 a \,b^{2}}}{b^{2}}\)	\(902\)
default	\(\text {Expression too large to display}\)	\(2152\)

[In]

int(x*(d*x^2+c)^(3/2)/(b*x^2+a)^2,x,method=_RETURNVERBOSE)

[Out]

3/2*(-d*(b*x^2+a)*(a*d-b*c)*arctan(b*(d*x^2+c)^(1/2)/((a*d-b*c)*b)^(1/2))+((a*d-b*c)*b)^(1/2)*(d*x^2+c)^(1/2)*
(1/3*(2*d*x^2-c)*b+a*d))/((a*d-b*c)*b)^(1/2)/b^2/(b*x^2+a)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 333, normalized size of antiderivative = 3.36 \[ \int \frac {x \left (c+d x^2\right )^{3/2}}{\left (a+b x^2\right )^2} \, dx=\left [\frac {3 \, {\left (b d x^{2} + a d\right )} \sqrt {\frac {b c - a d}{b}} \log \left (\frac {b^{2} d^{2} x^{4} + 8 \, b^{2} c^{2} - 8 \, a b c d + a^{2} d^{2} + 2 \, {\left (4 \, b^{2} c d - 3 \, a b d^{2}\right )} x^{2} - 4 \, {\left (b^{2} d x^{2} + 2 \, b^{2} c - a b d\right )} \sqrt {d x^{2} + c} \sqrt {\frac {b c - a d}{b}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) + 4 \, {\left (2 \, b d x^{2} - b c + 3 \, a d\right )} \sqrt {d x^{2} + c}}{8 \, {\left (b^{3} x^{2} + a b^{2}\right )}}, -\frac {3 \, {\left (b d x^{2} + a d\right )} \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {{\left (b d x^{2} + 2 \, b c - a d\right )} \sqrt {d x^{2} + c} \sqrt {-\frac {b c - a d}{b}}}{2 \, {\left (b c^{2} - a c d + {\left (b c d - a d^{2}\right )} x^{2}\right )}}\right ) - 2 \, {\left (2 \, b d x^{2} - b c + 3 \, a d\right )} \sqrt {d x^{2} + c}}{4 \, {\left (b^{3} x^{2} + a b^{2}\right )}}\right ] \]

[In]

integrate(x*(d*x^2+c)^(3/2)/(b*x^2+a)^2,x, algorithm="fricas")

[Out]

[1/8*(3*(b*d*x^2 + a*d)*sqrt((b*c - a*d)/b)*log((b^2*d^2*x^4 + 8*b^2*c^2 - 8*a*b*c*d + a^2*d^2 + 2*(4*b^2*c*d
- 3*a*b*d^2)*x^2 - 4*(b^2*d*x^2 + 2*b^2*c - a*b*d)*sqrt(d*x^2 + c)*sqrt((b*c - a*d)/b))/(b^2*x^4 + 2*a*b*x^2 +
a^2)) + 4*(2*b*d*x^2 - b*c + 3*a*d)*sqrt(d*x^2 + c))/(b^3*x^2 + a*b^2), -1/4*(3*(b*d*x^2 + a*d)*sqrt(-(b*c -
a*d)/b)*arctan(-1/2*(b*d*x^2 + 2*b*c - a*d)*sqrt(d*x^2 + c)*sqrt(-(b*c - a*d)/b)/(b*c^2 - a*c*d + (b*c*d - a*d
^2)*x^2)) - 2*(2*b*d*x^2 - b*c + 3*a*d)*sqrt(d*x^2 + c))/(b^3*x^2 + a*b^2)]

Sympy [F]

\[ \int \frac {x \left (c+d x^2\right )^{3/2}}{\left (a+b x^2\right )^2} \, dx=\int \frac {x \left (c + d x^{2}\right )^{\frac {3}{2}}}{\left (a + b x^{2}\right )^{2}}\, dx \]

[In]

integrate(x*(d*x**2+c)**(3/2)/(b*x**2+a)**2,x)

[Out]

Integral(x*(c + d*x**2)**(3/2)/(a + b*x**2)**2, x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {x \left (c+d x^2\right )^{3/2}}{\left (a+b x^2\right )^2} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(x*(d*x^2+c)^(3/2)/(b*x^2+a)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for
more detail

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.23 \[ \int \frac {x \left (c+d x^2\right )^{3/2}}{\left (a+b x^2\right )^2} \, dx=\frac {\sqrt {d x^{2} + c} d}{b^{2}} + \frac {3 \, {\left (b c d - a d^{2}\right )} \arctan \left (\frac {\sqrt {d x^{2} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{2 \, \sqrt {-b^{2} c + a b d} b^{2}} - \frac {\sqrt {d x^{2} + c} b c d - \sqrt {d x^{2} + c} a d^{2}}{2 \, {\left ({\left (d x^{2} + c\right )} b - b c + a d\right )} b^{2}} \]

[In]

integrate(x*(d*x^2+c)^(3/2)/(b*x^2+a)^2,x, algorithm="giac")

[Out]

sqrt(d*x^2 + c)*d/b^2 + 3/2*(b*c*d - a*d^2)*arctan(sqrt(d*x^2 + c)*b/sqrt(-b^2*c + a*b*d))/(sqrt(-b^2*c + a*b*
d)*b^2) - 1/2*(sqrt(d*x^2 + c)*b*c*d - sqrt(d*x^2 + c)*a*d^2)/(((d*x^2 + c)*b - b*c + a*d)*b^2)

Mupad [B] (verification not implemented)

Time = 5.51 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.18 \[ \int \frac {x \left (c+d x^2\right )^{3/2}}{\left (a+b x^2\right )^2} \, dx=\frac {\sqrt {d\,x^2+c}\,\left (\frac {a\,d^2}{2}-\frac {b\,c\,d}{2}\right )}{b^3\,\left (d\,x^2+c\right )-b^3\,c+a\,b^2\,d}+\frac {d\,\sqrt {d\,x^2+c}}{b^2}-\frac {3\,d\,\mathrm {atan}\left (\frac {\sqrt {b}\,d\,\sqrt {d\,x^2+c}\,\sqrt {a\,d-b\,c}}{a\,d^2-b\,c\,d}\right )\,\sqrt {a\,d-b\,c}}{2\,b^{5/2}} \]

[In]

int((x*(c + d*x^2)^(3/2))/(a + b*x^2)^2,x)

[Out]

((c + d*x^2)^(1/2)*((a*d^2)/2 - (b*c*d)/2))/(b^3*(c + d*x^2) - b^3*c + a*b^2*d) + (d*(c + d*x^2)^(1/2))/b^2 -
(3*d*atan((b^(1/2)*d*(c + d*x^2)^(1/2)*(a*d - b*c)^(1/2))/(a*d^2 - b*c*d))*(a*d - b*c)^(1/2))/(2*b^(5/2))